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 Some Squares Sum Square (Posted on 2012-07-30)
Determine all possible non zero real triplet(s) (F, G, H) that satisfy the following system of equations:

(7F)2 + (7G) 2 + (7H) 2 = (2F + 3G + 6H) 2 , and :

1/F + 1/G + 1/H = 1

 No Solution Yet Submitted by K Sengupta No Rating

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 Solution | Comment 6 of 9 |
This 3D approach is the only way I can seem to get my head round this problem and confirm that only one triple exists.

Let        (7F)2 + (7G)2 + (7H)2 = 49r2 ,      so that F2 + G2 + H2 = r2

then      (2F + 3G + 6H)2 = 49r2,  giving  2F + 3G + 6H = +/- 7r   (r>0 say)

Considering (F, G, H) to be coordinates in 3-D space...

The first equation is of a sphere with centre at (0, 0, 0) and radius r, while
the second equation is of a plane whose normal has direction ratios (2, 3, 6).

The distance of the plane from the origin is    +/- 7r/sqrt(22 + 32 + 62),

which simplifies to  +/- r, so the plane can only touch the sphere at the diametrically opposite points A (2r/7, 3r/7, 6r/7) and B (-2r/7, -3r/7, -6r/7).

Substitution into the second given equation, 1/F + 1/G + 1/H = 1, then shows that only A gives a possible solution, viz  r = 7/2 + 7/3 + 7/6 = 7, from which the only possible triple (F, G, H) is (2, 3, 6).

 Posted by Harry on 2012-08-01 13:12:16

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